When a cube of side length $a$ is in limiting equilibrium on an inclined plane and it is about to topple,what is the coefficient of friction between the cube and the plane?

$A$ block of mass $m$ is placed on a surface with a vertical cross-section given by $y = \frac{x^3}{6}$. If the coefficient of friction is $0.5$,the maximum height above the ground at which the block can be placed without slipping is:

$A$ small block starts slipping down from a point $B$ on an inclined plane $AB$,which is making an angle $\theta$ with the horizontal. The section $BC$ is smooth and the remaining section $CA$ is rough with a coefficient of friction $\mu$. It is found that the block comes to rest as it reaches the bottom (point $A$) of the inclined plane. If $BC = 2AC$,the coefficient of friction is given by $\mu = k \tan \theta$. The value of $k$ is

$A$ small mass slides down an inclined plane of angle $\theta$ with the horizontal. The coefficient of friction is $\mu = \mu_0 x$,where $x$ is the distance through which the mass slides down and $\mu_0$ is a constant. Then,the distance covered by the mass before it stops is:

$A$ block of mass $m$ is lying on an inclined plane. The coefficient of friction between the plane and the block is $\mu$. The force $(F_1)$ required to move the block up the inclined plane will be

$A$ block of mass $m$ is lying on an inclined plane. The coefficient of friction between the plane and the block is $\mu$. The force $(F)$ required to move the block up the inclined plane will be